Existence of a sequence and a positive number $\delta$, provided the given metric space is not totally bounded

Question

A few months ago, I came across the following question:-

Let $\left( X, d \right)$ be a metric space which is bounded but not totally bounded. Show that there is a positive number $\delta$ and a sequence $\left( x_n \right)$ such that for all $n \neq m$, $d \left( x_n, x_m \right) > \delta$.

I tried solving it by taking the negation of the statement,

$$\forall \delta > 0 \text{ and } \left( x_n \right), \exists m, n \text{ with } m \neq n \text{ such that } d \left( x_n, x_m \right) \leq \delta.$$

Then, I wished to prove that assuming this, the metric space would become totally bounded (which would contradict the hypothesis).

To prove that the metric space is totally bounded, I was thinking of proving that the sequence taken in the statement actually forms an delta-net for each positive delta. However, I am unsure how would that happen. This is because of two reasons:-

1. I do not know what the sequence actually is. The statement we have assumed is true for any sequence.

2. Even if we take a particular sequence (arbitrarily), how can we show that all the points of $X$ will be covered by that sequence using open balls of delta radius?

Any help will be appreciated!

Answer 1

Proof 1). There exist $\delta >0$ such the finite number of balls of radius $\delta$ cannot cover $X$. Pick any $x_1 \in X$. If $d(x,x_1) <\delta$ for all $x \in X$ the $X$ is covered by single ball of radius $delta$ which is a contradiction. So there exists $x_2$ such that $d(x_1,d_2) \geq \delta$. Now $B(x_1,\delta)$ and $B(x_2,\delta)$ do not cover $X$ so there exist $x_3$ such that $d(X_3,x_1) \geq \delta$ and $d(X_3,x_2) \geq \delta$ and so on. By induction we get the desired sequence $(x_n)$.

Proof 2): The negation of what is required is: for any $\delta >0$ there are at most finitely many points $\{x_1,x_2,...,x_n\}$ such that $d(x_i,x_j) \geq \delta $ for $i \neq j$. For $\delta =\frac 1 k$ let $N_k$ be the maximum cardinality of a set of this type. Then $X \subset \bigcup_{i=1}^{N_k} B(x_i,\frac 1 k)$ for suitable $x_i$'s. Given $\epsilon >0$ we can choose $k$ such that $\frac 1 k <\epsilon$ and this proves total boundedness.

PS: It is possible that for a certain $\delta$ there are no two points $x,y$ with $d(x,y) >\delta$ but that is fine here since the space will be covered by a single $\delta$ ball in this case.